4 Nonlinear equations solve for 4 variables
이전 댓글 표시
%script [ error- fsolve stopped because the problem appears to be locally singular.]
%solving E1,E2,E3,Noi
%Initial guesses
%E1o = 0.00009;
%E2o = 0.00009;
%E3o = 0.0009;
%Noio = 0.1;
X0 = [E1o; E2o; E3o; Noio];
%call fsolve function
%Call fsolve function
fhandle = @fsolve_function;
opts=optimoptions('fsolve','Display','iter','TolFun',1e-30,'TolX',1e-30);
[X,fval,exitflag] = fsolve (fhandle, X0,opts);
disp ('E1, E2, E3, Noi')
%equilibrium composition- function
function F=fsolve_function(X)
%unpack variables
E1=X(1);
E2=X(2);
E3=X(3);
Noi=X(4);%initial value of oxygen
K1=4.669E20;
K2=1.29;
K3=3.44E19;
yo2eq=0.00906;%atm
%Initial guesses
E1o = 0.00009;
E2o = 0.00009;
E3o = 0.0009;
Noio = 0.1;
%input equations
eq1=((2*E1-E2)^2/((Noi-E1)*(Noi+E1-E2+2*E3)))-K1;
eq2=(E2*(E2-2*E3)/(2*E1-E2)/(2*E3-E2))-K2;
eq3=((2*E3-E2)^2*(Noi+E1-E2+2*E3)/(E2-2*E3)^2/(Noi-E1))-K3;
eq4=((Noi-E1-E3)/(Noi+E1-E2+2*E3))-yo2eq;
F=[eq1; eq2; eq3; eq4; ];
end
답변 (1개)
%Initial guesses
E1o = 0.00009;
E2o = 0.00009;
E3o = 0.0009;
Noio = 0.1;
X0 = [E1o; E2o; E3o; Noio];
%call fsolve function
%Call fsolve function
fhandle = @fsolve_function;
opts=optimoptions('fsolve','Display','iter','TolFun',1e-30,'TolX',1e-30);
[X,fval,exitflag] = fsolve (fhandle, X0,opts)
disp ('E1, E2, E3, Noi')
%equilibrium composition- function
function F=fsolve_function(X)
%unpack variables
E1=X(1);
E2=X(2);
E3=X(3);
Noi=X(4);%initial value of oxygen
K1=4.669E20;
K2=1.29;
K3=3.44E19;
yo2eq=0.00906;%atm
%input equations
eq1=(2*E1-E2)^2-K1*(Noi-E1)*(Noi+E1-E2+2*E3);
eq2=E2*(E2-2*E3)-K2*(2*E1-E2)*(2*E3-E2);
eq3=(2*E3-E2)^2*(Noi+E1-E2+2*E3)-K3*(E2-2*E3)^2*(Noi-E1);
eq4=(Noi-E1-E3)-yo2eq*(Noi+E1-E2+2*E3);
F=[eq1; eq2; eq3; eq4; ];
end
댓글 수: 2
Mahe Rukh
2022년 9월 4일
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared and the vector of function values
is near zero as measured by the value of the function tolerance.
This is a message of success - fsolve found a solution.
How to solve this? and all the variables should have poisitve values.
If you need positive solutions, try different initial values or straight away "MultiStart" with the following code.
%Initial guesses
E1o = 0.00009;
E2o = 0.00009;
E3o = 0.0009;
Noio = 0.1;
X0 = [E1o; E2o; E3o; Noio];
X0 = sqrt(X0);
%Call fsolve function
fhandle = @fsolve_function;
opts=optimoptions('fsolve','Display','iter','TolFun',1e-30,'TolX',1e-30);
[X,fval,exitflag] = fsolve (fhandle, X0,opts)
norm(fsolve_function(X))
X = X.^2
%equilibrium composition- function
function F=fsolve_function(X)
%unpack variables
E1=X(1)^2;
E2=X(2)^2;
E3=X(3)^2;
Noi=X(4)^2;%initial value of oxygen
K1=4.669E20;
K2=1.29;
K3=3.44E19;
yo2eq=0.00906;%atm
%input equations
eq1=(2*E1-E2)^2-K1*(Noi-E1)*(Noi+E1-E2+2*E3);
eq2=E2*(E2-2*E3)-K2*(2*E1-E2)*(2*E3-E2);
eq3=(2*E3-E2)^2*(Noi+E1-E2+2*E3)-K3*(E2-2*E3)^2*(Noi-E1);
eq4=(Noi-E1-E3)-yo2eq*(Noi+E1-E2+2*E3);
F=[eq1; eq2; eq3; eq4; ];
end
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